Đáp án+Giải thích các bước giải:
$M=\dfrac{3}{4}+|x+\dfrac{1}{2}|$
$|x+\dfrac{1}{2}|≥0∀x$
$⇒\dfrac{3}{4}+|x+\dfrac{1}{2}|≥\dfrac{3}{4}$
Dấu $"="$ xảy ra khi
$x+\dfrac{1}{2}=0$
$⇒x=\dfrac{-1}{2}$
Vậy $M_{min}=\dfrac{3}{4}⇔x=\dfrac{-1}{2}$
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$N=-|4-x|+\dfrac{2}{3}$
$-|4-x|≤0∀x$
$⇒-|4-x|+\dfrac{2}{3}≤\dfrac{2}{3}$
Dấu $"="$ xảy ra khi
$4-x=0$
$⇒x=4$
Vậy $N_{max}=\dfrac{2}{3}⇔x=4$
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a, $|x-1,7|=2,3$
$⇒\left[\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.$
$⇒\left[\begin{matrix}x=4\\x=-0,6\end{matrix}\right.$
Vậy `x∈{4;-0,6}`
